Show that all the line segments drawn from a given point not on it the perpendicular line segment is the shortest
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◆Ekansh Nimbalkar◆
hello friend here is your required answer
hello friend here is your required answer
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Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM,
∠N = 90º
∠P + ∠N + ∠M = 180º (Angle sum property of a triangle)
∠P + ∠M = 90º
Clearly, ∠M is an acute angle.
∴ ∠M < ∠N
⇒ PN < PM (Side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM,
∠N = 90º
∠P + ∠N + ∠M = 180º (Angle sum property of a triangle)
∠P + ∠M = 90º
Clearly, ∠M is an acute angle.
∴ ∠M < ∠N
⇒ PN < PM (Side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
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